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A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 254 kcal of heat is added to the gas, the volume is observed to increase slowly from 12.0 m^3 to 16.2 m^3.
a. Calculate the work done by the gas.
b. Calculate the change in internal energy of the gas.



Answer :

The work done by the gas for question A will be 4.2 * 10^5 J.

The change in internal energy of the gas for question B is 6.43 * 10^5 J.

What is work (in gas)?

For a gas, it is the product of the pressure P and the volume V during a change of volume. The formula for work done in gas is W (work) = P (pressure) * V (volume)

Now, let's calculate the work done by gas to answer question A.

Work = Pressure * (Volume 2 - Volume 1)

= 1 atm * (16.2 - 12)

= 10^5 Pa * 4.2

= 4.2 * 10^5 J

So the work done by the gas is 4.2 * 10^5 J.

Now, let's find the change in internal energy of the gas.

The heat energy added to the system is

Q = 254kcal

= 254kcal * (4184J / 1 kcal)

= 1.063 * 10^6 J

The change in internal energy of gas is

ΔU = Q (the heat energy added) - W (work done by gas)

= 1.063 * 10^6 J - 4.2*10^5 J

= 6.43 * 10^5 J

Therefore, the change in internal energy of the gas is 6.43 * 10^5 J

Learn more about work (in gas) https://brainly.com/question/29589757

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